Answer:
b) - 3.3 J
Explanation:
Given;
mass, m = 2 kg
initial extension of the spring, x = 6 cm = 0.06 m
The weight of the mass on the spring;
W = mg
where;
g is acceleration due to gravity = 9.81 m/s²
W = 2 x 9.81
W = 19.62 N
The spring constant is calculated as;
W = kx
k = W/x
k = 19.62 / 0.06
k = 327 N/m
The work done by the spring when it is extended to an additional 10 cm;
work done = force x distance
distance = extension, x = 10 cm = 0.1 m
The work done by the spring opposes the applied force by acting in opposite direction to the force.
W = - Fx
W = - (kx) x
W = - kx²
W = - (327) x (0.1)²
W = - 3.27 J
W ≅ - 3.3 J
Therefore, the work done by the spring by opposing the applied force is -3.3 J
11. From this lab we can conclude that a) the heat transferred when objects are rubbed together creates an energy that can cause objects to move towards or away from each other. b) objects such as balloons and sweaters have a natural affinity towards each other. They will attract each other whether they are rubbed together or not. c) charges exert forces on other charges. do) charges do not exert forces on other charges.
Answer:
c) charges exert forces on other charges.
Explanation:
When two different materials are rubbed together, there is a transfer of electrons from one material to the other material so this causes one object to become positively charged and the other object is negatively charged so they will attract each other not repel each other. Charges exert forces on other charges i.e. opposite charges attract each other whereas similar charges repel each other so in both cases force are exerted on one another.
HELP ME ASAP PLSSS!!
what is meant by specific latent heat of vaporization of water is -2.26mjkg^-1 or -2.26mj/kg?
Answer:
The specific latent heat of a substance is the amount of energy required to change the state of one kilo of the substance without change in it temperature.The latent heat of vaporization or evaporation is the heat given to some mass to convert if from the liquid to the vapor phase.
A 2.0 kg frictionless puck is at rest on a level table. It is pushed straight north with a constant force of 5N for 1.50 s and then let go. How far does the puck move from rest in 2.5 s?
Answer:
the distance moved by the puck after 2.5 s is 7.8 m
Explanation:
Given;
mass of the puck, m = 2 kg
initial velocity of the puck, u = 0
applied force, F = 5 N
time of motion, t = 1.5 s
Acceleration of the puck is calculated from Newton's second law of motion;
F = ma
a = F/m
a = 5/2
a = 2.5 m/s²
The distance moved by the puck after 2.5 s is calculated as;
s = ut + ¹/₂at
s = 0 + ¹/₂at²
s = ¹/₂at²
s = 0.5 x 2.5 x (2.5)²
s = 7.8 m
Therefore, the distance moved by the puck after 2.5 s is 7.8 m
define nortons theorem
Answer:
In direct-current circuit theory, Norton's theorem is a simplification that can be applied to networks made of linear time-invariant resistances, voltage sources, and current sources. At a pair of terminals of the network, it can be replaced by a current source and a single resistor in parallel.
what is the distance time how can we find the speed of an object from its distance time graph
Answer:
speed is the gradient of the graph
Answer:
Speed is the slope of a distance time graph.
Explanation:
Speed= d/t
Slope is equal to rise/run
If the rise of the graph is the distance and the run is the time, calculating slope is the equivalent of calculating average speed.
an object of volume has 20ml has a mass of 2.5kg what will be its density
Answer:
0.125
Explanation:
Density =mass/volume
Density =2.5/20
Density =0.125
Answer:
D=m/v
=2.5kg/(20/1000000)m^3
2.5kg÷0.000002m^3
1250000kgm^-3
Explanation:
20 is divided by 1000000 because m^3 is its si unit
) The velocity function is v(t)=−t2+3t−2v(t)=−t2+3t−2 for a particle moving along a line. Find the displacement (net distance covered) of the particle during the time interval [−2,5][−2,5].
Answer:
89.87m/s
Explanation:
Given the velocity function
v(t)=−t²+3t−2
In order to get the displacement function, we will integrate the velocity function as shown:
[tex]\int\limits^5_{-2} {v(t)} \, dt \\d(t)= \int\limits^5_{-2}{(-t^2+3t+2)} \, dt \\\\d(t)=[\frac{-t^3}{3}+\frac{3t^2}{2}+2t ]^5_{-2}\\[/tex]
at t = 5
[tex]d(5)=[\frac{-5^3}{3}+\frac{3(5)^2}{2}+2(5) ]\\d(5)=[\frac{-125}{3}+\frac{75}{2}+10 ]\\d(5)=-41.7+37.5+10\\d(5)=89.2m/s[/tex]
at t = -2
[tex]d(-2)=[\frac{-(-2)^3}{3}+\frac{3(-2)^2}{2}+2(-2) ]\\d(-2)=[\frac{-8}{3}+\frac{12}{2}+(-4) ]\\d(-2)=-2.67+6-4\\d(-2)=-0.67m/s[/tex]
Required displacement = d(5) - d(-2)
Required displacement = 89.2 - (-0.67)
Required displacement = 89.2 + 0.67
Required displacement = 89.87m/s
What do you understand by moment of inertia and torque?
Word limit 50-60
Please don't copy from any sources. You can rewrite. Plagiarism will be check. Thank you.
Answer:
Moment of inertia, in physics, quantitative measure of the rotational inertia of a body—i.e., the opposition that the body exhibits to having its speed of rotation about an axis altered by the application of a torque (turning force). The axis may be internal or external and may or may not be fixed.
Which of the following statements about magnetism is TRUE?
a) The direction of the magnetic force on a current-carrying wire is parallel to the wire.
b) Magnetic poles always occur in pairs (N and S).
c) Magnetic field lines begin at south poles and end on north poles.
d) Moving charges do not experience a force in magnetic fields.
Two long, straight wires are separated by 0.120 m. The wires carry currents of 11 A in opposite directions, as the drawing indicates. Find the magnitude of the net magnetic field.
Answer:
The magnitude of the magnetic field is 1.83 x [tex]10^{-5}[/tex] T.
Explanation:
The flow of an electric current in a straight wire induces magnetic field around the wire. When current is flowing through two wires in the same direction, a force of attraction exists between the wires. But if the current flows in opposite directions, the force of repulsion is felt by the wires.
In the given question, the direction of flow of current through the wires is opposite, thus both wires applies the same field on each other. The result to repulsion between them.
The magnetic field (B) between the given wires can be determined by:
B = [tex]\frac{U_{o}I }{2\pi r}[/tex]
where: I is the current, r is the distance between the wires and [tex]U_{0}[/tex] is the magnetic field constant.
But, I = 11 A, r = 0.12 m and [tex]U_{0}[/tex] = 4[tex]\pi[/tex] x [tex]10^{-7}[/tex] Tm/A
So that;
B = [tex]\frac{4\pi *10^{-7}*11 }{2\pi *0.12}[/tex]
= 1.8333 x [tex]10^{-5}[/tex]
B = 1.83 x [tex]10^{-5}[/tex] T
Given that the temperature of a body is 527K determine the value in degree C
Answer:
253.85°C
Explanation:
Here is the formula for converting K to °C
527K − 273.15 = 253.85°C
If the accuracy in measuring the velocity of a particle increases, the accuracy in measuring its position will:__________.
a. It is impossible to say since the two measurements are independent and do not affect each other.
b. remain the same.
c. increase.
d. decrease.
When the accuracy in measuring the velocity of a particle increases, that of its position decreases. The correct option is d. decrease.
The Heisenberg's uncertainty principle states that there is always an uncertainty in any attempt to measure accurately the value of any two complementary variables simultaneously. This implies that measuring of position and momentum of a particle simultaneously would lead to uncertainty in their values.
So that;
Δx.ΔP ≥ [tex]\frac{h}{2\pi }[/tex]
Where Δx is the uncertainty in the value of its position, ΔP is the uncertainty in the value of momentum and h is the Planck's constant.
This principle simply explains and provides the required answer to the given question. So that according to Heisenberg's uncertainty principle, the accuracy in measuring its position will definitely decrease.
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The velocity of a body is given by the equation v=a+bx, where 'x' is displacement. The unit of b is
Answer:
2bsnsnsnns181991oiwiw
Several years ago, there was a solar eclipse visible only from a remote part of Mexico (Los Cabos). To view this event, many thousands of tourists went to that remote area. What effect would this event have on the price of pesos in tersm of U.S. dollars
The price of Pesos is increased, and it takes more to buy a given amount of Pesos
According to the demand and supply economic theory, made popular by Adam Smith in 1776, which provides a price determination model based on the relationship between the amount of a good producers of the good are willing to sell at a given price and the amount of that good that the end users wish to purchase, there is an inverse relationship between price and supply, therefore as the supply decreases, the price increases, and therefore, there is a direct relationship between the demand and price, such that as the demand increases, the price of the goods increases because the ratio of supply to demand is decreased
Given that the arrival of thousands of tourists to the remote Los Cabos, will result in the tourists requiring Pesos, the demand of Pesos will rise, which according to the demand and supply theory, will increase the price of Pesos
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A plastic dowel has a Young's Modulus of 1.50 ✕ 1010 N/m2. Assume the dowel will break if more than 1.50 ✕ 108 N/m2 is exerted. What is the maximum force (in kN) that can be applied to the dowel assuming a diameter of 2.30 cm?
A.
52.3 kN
B.
62.3 kN
C.
72.3 kN
D.
42.3 N
Answer:
cobina
Explanation:
me 2
when do things move faster? Day or Night?
Suppose a teenager on her bicycle. The rear wheel is spinning at an angular velocity of 281.133 rpm. She stops it in 3.686 s. How many revolution did it take to stop it?
Answer:
Explanation:
The formula for angular velocity is
[tex]\omega=\frac{\theta}{t}[/tex] where omega is the angular velocity, theta is the change in the angular rotation, and t is the time in seconds. First and foremost, we have the angular rotation in minutes and the time in seconds, so that's a problem we have to amend. Let's change the angular rotation to rotations per second:
[tex]281.133\frac{r}{min}*\frac{1min}{60s}=4.68555\frac{r}{s}[/tex]
Now we're ready to set up the problem:
[tex]4.68555=\frac{\theta}{3.686}[/tex] and we multiply both sides by 3.686 to get the rotations per seconds:
θ = 17.27 rotations
If a wave has to travel 600m and it’s wavelength is 0.4 m , with a frequency of 500Hz. How much time will it take for the wave to travel 600m ?
If Light is travels from air into pure water with an incident angle of 30°. What is the angle of refraction?
Explanation:
For air, n1 = 1.00003; for water, n2 = 1.3330
Given: θ2 = 30 degrees, then
θ1 = arcsin [(n2/n1) sin θ2]
= arcsin [(1.3330/1.0003) sin (40)]
= 58.93 degrees
Note that since, in this example, light is traveling from a medium of higher density (water; n2 = 1.3330) to a medium of lower density (air; n1 = 1.0003), then n2 > n1, and the angle of refraction (θ1) is larger than the angle of incidence (θ2), thus the light bends away from the normal (in this example, the vertical) as it leaves the water and enters the air.
(a) Calculate the force needed to bring a 800 kg car to rest from a speed of 85.0 km/h in a distance of 115 m (a fairly typical distance for a nonpanic stop).
N
(b) Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m. Calculate the force exerted on the car and compare it with the force found in part (a), i.e. find the ratio of the force in part(b) to the force in part(a).
(force in part (b) / force in part (a))
Answer:
2Al + 2H2O + 2NaOH ⟶ 3H2 + 2NaAlO2
Chất rắn màu xám bạc của nhôm (Al) tan dần trong dung dịch, sủi bọt khí là hidro (H2).
Explanation:
A particle moves along X-axis in such a way that X-coordinate varies with time according to expression x= 2-5t+6t2 meters, Calculate the initial velocity of the particle?
A 5
v= dt/ dx =−5+12t
Initial velocity means at t=0, which is −5+0=−5.
Thus, −v=5n
A 2000 kg truck has put its front bumper against the rear bumper of a 2500 kg SUV to give it a push. With the engine at full power and good tires on good pavement, the maximum forward force on the truck is 18,000 N.
What is the maximum possible acceleration the truck can give the SUV?
At this acceleration, what is the force of the SUV's bumper on the truck's bumper?
Answer:
The net magnitude of the force of the SUV's bumper on the truck's bumper is 9120 N.
Explanation:
Concepts and reason
The concept required to solve this problem is Newton’s second law of motion.
Initially, write an expression for the force according to the Newton’s second law of motion. Later, rearrange the expression for the acceleration. Finally, substitute the value of the acceleration obtained to find the new force.
Fundamentals
According to the Newton’s second law of motion, the net force is equal to the product of the mass and the acceleration of an object. The expression for the Newton’s second law of motion is as follows:
F = maF=ma
Here, m is mass and a is the acceleration.
(a)
Rearrange the equation F = maF=ma for a.
a = \frac{F}{m}a=
m
F
Substitute 18,000 N for F and \left( {2300{\rm{ kg + 2400 kg}}} \right)(2300kg+2400kg) for m in the equation a = \frac{F}{m}a=
m
F
.
\begin{array}{c}\\a = \frac{{18,000{\rm{ N}}}}{{\left( {2300{\rm{ kg + 2400 kg}}} \right)}}\\\\ = \frac{{18,000{\rm{ N}}}}{{\left( {4700{\rm{ kg}}} \right)}}\\\\ = 3.83{\rm{ m/}}{{\rm{s}}^2}\\\end{array}
a=
(2300kg+2400kg)
18,000N
=
(4700kg)
18,000N
=3.83m/s
2
(b)
Substitute 3.83{\rm{ m/}}{{\rm{s}}^2}3.83m/s
2
for a and 2400 kg for m in the equation F = maF=ma .
\begin{array}{c}\\F = \left( {2400{\rm{ kg}}} \right)\left( {3.83{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 9120{\rm{ N}}\\\end{array}
F=(2400kg)(3.83m/s
2
)
=9120N
Ans: Part a
The maximum possible acceleration the truck can give the SUV is 3.83{\rm{ m/}}{{\rm{s}}^2}3.83m/s
2
.
Part b
The net magnitude of the force of the SUV's bumper on the truck's bumper is 9120 N.
The maximum possible acceleration the truck can give the SUV is equal to 4 m/s².
The force of the SUV's bumper on the truck's bumper is 10000N
What is acceleration?Acceleration of an object can be described as as the change in the velocity of an object w.r.t. time. The acceleration is a vector quantity, contains both magnitude and direction. Acceleration is the second derivative of position w.r.t. time and the first derivative of velocity w.r.t. time.
According to Newton's second law of motion, the force is equal to the product of acceleration and mass of an object.
F = ma
And, a = F/m
Given, the mass of the ruck , m = 2000 Kg
The mass of the SUV, M = 2500 Kg
The total mass of the both = 2000 + 2500 = 4500 Kg
The maximum force on the trick , F = 18000 N
The maximum acceleration of the truck can give the SUV:
[tex]a_{max} = \frac{F_{max}}{m+M}[/tex]
a = 18000/4500
a = 4 m/s²
The force of the SUV's bumper on the truck's bumper will be:
[tex]F_{max} -f= ma_{max}[/tex]
[tex]f= 18000-2000\times 4[/tex]
[tex]f =10000N[/tex]
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Convert 385k to temperature of
Answer:
233.33°F
Explanation:
(385K - 273.15) * 9/5 + 32 = 233.33°F
An electron is released from rest at a distance of 9.00 cm from a fixed proton. How fast will the electron be moving when it is 3.00 cm from the proton
Answer:
the speed of the electron at the given position is 106.2 m/s
Explanation:
Given;
initial position of the electron, r = 9 cm = 0.09 m
final position of the electron, r₂ = 3 cm = 0.03 m
let the speed of the electron at the given position = v
The initial potential energy of the electron is calculated as;
[tex]U_i = Fr = \frac{kq^2}{r^2} \times r = \frac{kq^2}{r} \\\\U_i = \frac{(9\times 10^9)(1.602\times 10^{-19})^2}{0.09} \\\\U_i = 2.566 \times 10^{-27} \ J[/tex]
When the electron is 3 cm from the proton, the final potential energy of the electron is calculated as;
[tex]U_f = \frac{kq^2}{r_2} \\\\U_f = [\frac{(9\times 10^9)\times (1.602 \times 10^{-19})^2}{0.03} ]\\\\U_f = 7.669 \times 10^{-27} \ J \\\\\Delta U = U_f -U_i\\\\\Delta U = (7.699\times 10^{-27} \ J ) - (2.566 \times 10^{-27} \ J)\\\\\Delta U = 5.133 \times 10^{-27} \ J[/tex]
Apply the principle of conservation of energy;
ΔK.E = ΔU
[tex]K.E_f -K.E_i = \Delta U\\\\initial \ velocity \ of \ the \ electron = 0\\\\K.E_f - 0 = \Delta U\\\\K.E_f = \Delta U\\\\\frac{1}{2} mv^2 = \Delta U\\\\where;\\\\m \ is \ the \ mass \ of\ the \ electron = 9.1 1 \times 10^{-31} \ kg\\\\v^2 = \frac{ 2 \Delta U}{m} \\\\v = \sqrt{\frac{ 2 \Delta U}{m}} \\\\v = \sqrt{\frac{ 2 (5.133\times 10^{-27})}{9.11\times 10^{-31}}}\\\\v = \sqrt{11268.935} \\\\v = 106.2 \ m/s[/tex]
Therefore, the speed of the electron at the given position is 106.2 m/s
Solve numerical problem. Please give me step - step explanation Help me out plz
Answer:
You should multiply 60 kg*9.8 and answer will come.
Hope this will help you.
Answer:
yes she is right you should multiple 60*9.8
have a great day God bless you
Consider a 200-ft-high, 1200-ft-wide dam filled to capacity. Determine (a) the hydrostatic force on the dam and (b) the force per unit area of the dam near the top and near the bottom. Note: we will see that the resultant hydrostatic force will be
Answer:
a) [tex]F_g=1.5*10^9Ibf[/tex]
b) [tex]F_t=12490Ibf/ft^2[/tex]
[tex]F_b=0[/tex]
Explanation:
From the question we are told that:
Height [tex]h=200ft[/tex]
Width [tex]w=1200ft[/tex]
a)
Generally the equation for Dam's Hydro static force is mathematically given by
[tex]F_g=\rho*g*\frac{h}{2}(w*h)[/tex]
Where
[tex]\rho=Density\ of\ water[/tex]
[tex]\rho=62.4Ibm/ft^3[/tex]
Therefore
[tex]F_g=62.4*32.2*\frac{200}{2}(1200*200)[/tex]
[tex]F_g=1.5*10^9Ibf[/tex]
b)
Generally the equation for Dam's Force per unit area is mathematically given by
[tex]F=\rho*g*h[/tex]
For Top
[tex]F_t=\rho*g*h[/tex]
[tex]F_t=62.4*32.2*200[/tex]
[tex]F_t=12490Ibf/ft^2[/tex]
For bottom
[tex]Here \\H=0 zero[/tex]
Therefore
[tex]F_b=0[/tex]
The hydrostatic force on the dam is [tex]2.995 \times 10^9 \ lbF[/tex].
The force per unit area near the top is 86.74 psi.
The force per unit area near the bottom is zero.
Hydrostatic force
The hydrostatic force on the dam is the force exerted on the dam by the column of the water.
[tex]F = PA\\\\F = (\rho gh) \times (wh)\\\\F = (62.4 \times 32.17 \times 200) \times (1200 \times 200)\\\\F = 9.636 \times 10^{10} \ lb-ft/s^2\\\\1 \ lbF = 32.17\ lb-ft/s^2\\\\F = 2.995 \times 10^9 \ lbF[/tex]
Force per unit area near the topThe force per unit area is the pressure exerted near the top of the dam.
[tex]P = \rho gh\\\\P = 0.052 \times \rho h[/tex]
where;
P is pressure in PSI
ρ is density of water in lb/gal
h is the vertical height in ft
[tex]P = 0.052 \times 8.34 \times 200\\\\P = 86.74 \ Psi[/tex]
The pressure near the bottom is zero, become the vertical height is zero.
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What is the name of the compound br8P4
Answer:
Octabromine tetraphosphide
Explanation:
This compound has in its formula:
- Eight bromines
- Four phosphorous
8 → octa prefix
4 → tetra prefix
Right answer is Octabromine tetraphosphide
Imagine that you and your lab partner are standing on smooth ice holding onto opposite ends of a long rope. There is no friction between your feet and the ice. You tug gently on the rope. (a)Describe the motion of you and your lab partner before, during, and after the tug. Be as specific as you can. Expla
Answer:
Following are the complete solution to the given question:
Explanation:
In this question, before tug they are at rest, thus their initial momentum is 0 [tex]\pi=0[/tex], according to the law of conservation [tex]p_f=0[/tex] thus both should move in the opposite direction whose sum of the momentums is equaled to zero as there is no friction they will continue to move with their velocities until their collision. Its collision doesn't happen it will continue up to an infinite distance.
The motion of both you and your partner depends on the tension in the rope.
We must note that frictional force is the force that acts between two surfaces in contact with each other. The frictional force depends on the nature of the surfaces in contact.
Now, since there is no friction between your leg and the ice. The motion of both you and your partner depends on the tension in the rope.
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An object undergoing simple harmonic motion takes 0.40 s to travel from one point of zero velocity to the next such point. The distance between those points is 50 cm. Calculate (a) the period, (b) the frequency, and (c) the amplitude of the motion.
Answer:
a) [tex]P=0.80[/tex]
b) [tex]1.25Hz[/tex]
c) [tex]A=25cm[/tex]
Explanation:
From the question we are told that:
Travel Time [tex]T=0.40s[/tex]
Distance [tex]d=50cm[/tex]
a)
Period
Time taken to complete one oscillation
Therefore
[tex]P=2*T\\\\P=2*0.40[/tex]
[tex]P=0.80[/tex]
b)
Frequency is
[tex]F=\frac{1}{T}\\\\F=\frac{1}{0.80}[/tex]
[tex]1.25Hz[/tex]
c)
Amplitude:the distance between the mean and extreme position
[tex]A=\frac{50}{2}[/tex]
[tex]A=25cm[/tex]
